Week 7 of 52: Conversion of alcohols to alkyl halides by hydrohalic acids

Yes, I'm still a week behind schedule. I know. Someday I'll even catch up. But not yet. Last time, I introduced a specific variation on elimination. So now let's try one for substitution. Are you excited? I know I am. This reaction is pretty fantastic, but it might not be what you're used to. Instead of using the properties of nucleophilic substitution to replace a halogen with something else, we're going to replace something else with a halogen. It's backwards!

As you almost certainly do not recall, halogens make good leaving groups and hydroxide makes a good nucleophile. So this really does seem backwards. How can we have a nucleophilic substitution reaction in which something that is ordinarily a good nucleophile is the leaving group and something that is ordinarily a good leaving group is the nucleophile. The answer, of course, is that we cheat. Come on, isn't that obvious? What might not be obvious is just how we are going to go about cheating. No, I'm just kidding. That's obvious too. No?

Fine. Remember how we can dehydrate alcohols? I mean, the last two reactions have been about that. We turn that bad leaving group into a good one. So here's your first hint: we'll turn that hydroxyl group into a good leaving group by protonating it with a strong acid. You get it now, right? No. Here's another hint: the title of this post mentions hydrohalic acids. That's right, hydrochloric, hydrobromic, hydroiodic. HCl, HBr, HI.

Do I really still have to spell it out for you? The acid protonates the oxygen, creating a good leaving group, then the halide attacks the molecule as a nucleophile. This occurs by an S_N2 mechanism for primary alcohols and an S_N1 mechanism for secondary and tertiary alcohols, of course. Isn't it great?

Also, bromide and iodide are strong enough nucleophiles for this, but chloride can require a catalyst for the reaction to progress. But no more about that for now.